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In the first answer to this MO post the author says that the $H^2$ of $X$ can be compute using the Cech-to-derived functor spectral sequence, i.e. in that case the Mayer-Vietoris sequence.

I'm having trouble in understanding how it works: as for $H^1(U\cap V)$, this should be equal to $$H^1_{et}(\operatorname{Spec} k((t)), A)=Cont(Gal(k((t))^s,k((t))),\mathbb Z/n)=Cont((1),\mathbb Z/n)\cong\mathbb Z/n.$$ (See the answer to this post, second part.) Is this correct?

But when I come to compute $H^1_{et}(U,A)$ and $H^1_{et}(V,A)$, which I would like to be $0$ in order to conclude that $H^2_{et}(X,A)\cong H^1_{et}(U\cap V,A)\cong \mathbb Z/n$, I am stuck. Can someone give me a clue?

Thank you in advance.

W.Rether
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    $k[[t]]$ is a strictly henselian ring, so there is no non-trivial étale covering. In other word, this is a point for the étale topology. Hence, it has vanishing higher cohomology. – Roland Jan 07 '19 at 18:36
  • Ok, thanks. There is just onething I'm missing: why is the same false for $k((t))$? – W.Rether Jan 07 '19 at 18:57
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    Because $k((t))$ is not algebraically closed. As you wrote, its $H^1$ is the group of continuous homomorphism from $\operatorname{Gal}(k((t))^s/k((t)))$ to $A$ (not $\mathbb{Z}/n\mathbb{Z}$). A classical picture : $k[[t]]$ is a small disc and $k((t))$ is the punctured disc. So $k[[t]]$ is contractible and it has no cohomology, $k((t))$ is homotopy equivalent to a circle so it has a $H^1\simeq A$. – Roland Jan 07 '19 at 19:30
  • Ok, thanks so much! In the example $A$ could be the constant sheaf $\mathbb Z/n$, I think, isn't it? (He says "of finite order"...). – W.Rether Jan 07 '19 at 19:55
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    Sure it is possible, but you didn't specify $A$. This is just for the formula $Cont(Gal,A)\simeq\mathbb{Z}/n\mathbb{Z}$ (which is false if $A\neq\mathbb{Z}/n\mathbb{Z}$). I just have a doubt : I believe, one need that $k$ is of characteristic $0$ or that $A$ has torsion prime to the characteristic. – Roland Jan 07 '19 at 20:01
  • Yes. He says "finite of order invertible in $A$", but I think that he misspelled for $k$. – W.Rether Jan 07 '19 at 20:24
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    This is because the Galois group of $k((t))$ as a huge $p$-torsion part. If $A$ has no $p$-torsion, then $Cont(G,A)$ factor through the "prime to $p$" quotient which is $\mathbb{Z}'(1)=\prod_{l\neq p}\mathbb{Z}_l(1)$. But if $A$ has $p$ torsion, this may be very different. – Roland Jan 07 '19 at 20:30

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